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A271665
Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that w^2 + 4*x*y + 8*y*z + 32*z*x is a square, where w is a positive integer and x,y,z are nonnegative integers.
38
1, 3, 1, 1, 6, 3, 1, 3, 1, 6, 2, 1, 7, 10, 1, 1, 9, 3, 2, 6, 2, 2, 3, 3, 8, 10, 1, 1, 10, 2, 2, 3, 5, 8, 11, 1, 7, 13, 2, 6, 16, 6, 1, 2, 6, 2, 3, 1, 3, 16, 4, 7, 9, 3, 2, 10, 4, 9, 4, 1, 8, 15, 1, 1, 15, 5, 2, 9, 6, 8, 2, 3, 10, 13, 4, 2, 17, 7, 1, 6
OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*3^m, 4^k*3^m*43, 4^k*9^m*q (k,m = 0, 1, 2, ... and q = 7, 15, 79, 95, 141, 159, 183).
(ii) Any positive integer n can be written as w^2 + x^2 + y^2 + z^2 with w*x + x*y + 2*y*z + 3*z*x (or w*x + 3*x*y + 8*y*z + 5*z*x) twice a square, where w is a positive integer and x,y,z are nonnegative integers.
(iii) For each k = 1, 2, 8, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with w^2 + k*(x*y+y*z) a square, where w is a positive integer and x,y,z are nonnegative integers.
(iv) For each ordered pair (b,c) = (16,4), (24,4), (32,16), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^2 + b*y^2 + c*x*z + c*y*z + c*z*w is a square.
We also guess that for each triple (b,c,d) = (1,3,4), (1,6,8), (1,7,24), (1,8,15), (1,10,24), (1,12,35), (1,14,48), (1,20,48), (2,1,2), (2,4,2), (2,4,7), (2,6,7), (2,8,4), (2,8,14), (2,8,31), (2,10,23), (2,12,14), (2,12,34), (2,14,47), (3,1,1), (3,1,4), (3,1,16), (3,2,14), (3,2,17), (3,2,38), (3,3,3), (3,3,13), (3,4,1), (3,4,4), (3,5,11), (3,6,3), (3,6,6), (3,6,26), (3,8,2), (3,8,13), (3,8,22), (3,9,39), (3,12,12), (3,12,33), (3,15,1), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and x^2 + b*y^2 + c*x*z + d*y*z a square.
See also A271510, A271513, A271518 and A271644 for other conjectures refining Lagrange's four-square theorem.
LINKS
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.
EXAMPLE
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1^2 + 4*0*1 + 8*1*1 + 32*1*0 = 3^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 4*2*0 + 8*0*0 + 32*0*0 = 2^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1^2 + 4*2*1 + 8*1*1 + 32*1*2 = 9^2.
a(15) = 1 since 15 = 1^2 + 2^2 + 1^2 + 3^2 with 1^2 + 4*2*1 + 8*1*3 + 32*3*2 = 15^2.
a(43) = 1 since 43 = 3^2 + 3^2 + 4^2 + 3^2 with 3^2 + 4*3*4 + 8*4*3 + 32*3*3 = 21^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5^2 + 4*3*6 + 8*6*3 + 32*3*3 = 23^2.
a(95) = 1 since 95 = 5^2 + 6^2 + 5^2 + 3^2 with 5^2 + 4*6*5 + 8*5*3 + 32*3*6 = 29^2.
a(129) = 1 since 129 = 5^2 + 6^2 + 8^2 + 2^2 with 5^2 + 4*6*8 + 8*8*2 + 32*2*6 = 27^2.
a(141) = 1 since 141 = 8^2 + 5^2 + 4^2 + 6^2 with 8^2 + 4*5*4 + 8*4*6 + 32*6*5 = 36^2.
a(159) = 1 since 159 = 11^2 + 1^2 + 6^2 + 1^2 with 11^2 + 4*1*6 + 8*6*1 + 32*1*1 = 15^2.
a(183) = 1 since 183 = 1^2 + 9^2 + 10^2 + 1^2 with 1^2 + 4*9*10 + 8*10*1 + 32*1*9 = 27^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x*y+8*y*z+32*z*x+(n-x^2-y^2-z^2)], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 0, Sqrt[n-1-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 12 2016
STATUS
approved